If we have an $n \times n$ matrix, with all entries taken modulo $p$, how can we output the three matrixes in LDU decomposition, with all entries again modulo $p$? We can assume the input matrix is invertible.
That is, $LDU=A$, with $A$ given. $L$ is a lower-triangular matrix, $D$ is a diagonal matrix, and $U$ is an upper-triangular matrix. The entries of the results are modulo $p$.
For reference, here is how to generate a modular $\mathbf L\mathbf D\mathbf U$ decomposition, as suggested by Daniel in the comments:
mat = {{1, 0, 2}, {2, 1, 3}, {2, 1, 2}}; m = 5;
{lu, piv, cond} = LUDecomposition[mat, Modulus -> m];
l = LowerTriangularize[lu, -1] + IdentityMatrix[Length[lu]];
d = DiagonalMatrix[Diagonal[lu]];
u = Mod[DiagonalMatrix[PowerMod[Diagonal[lu], -1, m]].UpperTriangularize[lu], m];
You can check that Mod[l.d.u, m] gives a result that is congruent to mat. If the matrix has a singular leading submatrix, one should then account for an additional permutation matrix constructed from the vector piv; how to do this is left as an exercise for the interested reader.
LUDecomposition[mat,Modulus->p]and separate out the diagonal from the upper part. $\endgroup$LUDecomposition[]pivots, so you have an extra permutation matrix to contend with. If you need the version without pivoting, you'll have to write your own. $\endgroup$LUDecomposition[{{4, 4, 4, 3}, {0, 0, 4, 1}, {3, 3, 4, 2}, {4, 1, 0, 4}}, Modulus -> 5]But that would be because the leading $2\times 2$ block is singular, which is exactly when pivoting is necessary. $\endgroup$